Could you help me prove the following fact? I've been trying to prove it and I've searched for a hint in Englking's book, but I haven't come up with anything:
If $X$ is a Hausdorff space and each $x \in X$ has at least one compact neighbourhood, then $X$ is locally compact.
$U$ is a neighbourhood of $x \in X$ $\iff$ $\exists V$ - open $: x \in V \subset U$.
$X$ is locally compact $\iff$ each $x \in X$ has a basis of compact neighbourhoods.
On
A tricky one, indeed. I already had that once. Here we go :
We suppose that every point of $X$ has at least one compact neighborhood.
Let $x\in X$ and let $V$ be an open neighborhood of $x$. We want to show that there exists a compact neighborhood $W$ of $x$ with $W \subset V$.
By hypothesis, $x$ has a compact neighborhood, let's call it $C$. Let's take the subspace topology for $C$, so we have that $C$ is a compact Hausdorff space.
Let's now take $U = C \cap V$. And now we have that in $C$, $U$ is an open neighborhood of $x$. We know that $C$ is locally compact since it's compact and Hausdorff.
Thus, in $C$ we have a compact neighborhood $W$ of $x$ with $W \subset U$, and then also $W \subset V$.
In the end, we must check that $W$ is a neighborhood of $x$ in $X$, too. But since a neighborhood in a neighborhood is a neighborhood, we have it directly.
Actually, a stronger result is true: If each point has a compact Hausdorff neighborhood, then $X$ is locally compact.
Let $x\in X$, $U$ a neighborhood of $x$, and $K$ a compact Hausdorff neighborhood of $x$. Since $K$ is regular as a subspace of $X$ and $U\cap K$ is a neighborhood of $x$ in $K,$ there is a neighborhood $C\subseteq U\cap K$ of $x$ in $K$ (and thus also in $X$ since $K$ is a neighborhood of $x$) which is closed in $K$, thus compact. This proves that there are arbitrary small compact neighborhoods around $x$.